Equations with no powers and no products, solved all together: what a solution is, why there are only three possible outcomes, and how a system becomes an augmented matrix that row operations can simplify.
A cafรฉ sells a coffee and a pastry. On Monday a customer pays 5 for two coffees and one pastry; on Tuesday another pays 1 more for a coffee than for a pastry. What does each item cost? Writing $x$ for the price of a coffee and $y$ for a pastry gives two equations, $2x+y=5$ and $x-y=1$, and the question is: which pair $(x,y)$ makes both true at once? Questions of exactly this shape โ several unknowns, several conditions, every condition a sum of "number $\times$ unknown" โ appear everywhere: in balancing the forces at the joints of a bridge, in mixing chemicals to a required concentration, in pricing goods across a whole economy, in finding a phone's position from satellite signals.
This section sets up the language for all of them. We say precisely what a linear equation and a system of them is, what counts as a solution, and why a system can only ever have no solution, exactly one, or infinitely many. Then we strip a system down to its numbers โ an augmented matrix โ and meet the three elementary row operations that change a system without changing its solutions. Section 1.2 turns those operations into a complete algorithm; here we learn what they are and why they are safe.
Linear Equations and Their Solutions
๐ Definition 1.1 โ Linear equation
A linear equation in the variables $x_1,x_2,\dots,x_n$ is an equation that can be written in the form
$$a_1x_1+a_2x_2+\cdots+a_nx_n=b,$$
where the coefficients $a_1,\dots,a_n$ and the constant term $b$ are real numbers. With two or three variables we usually write $x,y$ or $x,y,z$ instead of $x_1,x_2,x_3$.
So a linear equation contains each variable only to the first power, multiplied by a constant. No products of variables ($xy$), no powers ($x^2$, $\sqrt x$), no variables inside other functions ($\sin x$, $e^y$, $\tfrac1x$). The coefficients themselves may be any real numbers โ $\sqrt2$, $\pi$ or $\sin\tfrac\pi6$ are fine, because they are just numbers.
โ๏ธ Example 1 โ Linear or not?
Decide which equations are linear: (a) $3x-2y+z=7$; (b) $x_1-4x_2+\sqrt2\,x_3=\pi$; (c) $xy+z=1$; (d) $\sqrt x+y=4$; (e) $\dfrac1x+y=2$; (f) $\left(\sin\tfrac\pi6\right)x+3y=0$.
๐ก Show solutionโถ
Linear (a) is already in the form of Definition 1.1 with $a_1=3$, $a_2=-2$, $a_3=1$, $b=7$. (b) is linear too: $\sqrt2$ and $\pi$ are constants. (f) is linear: $\sin\tfrac\pi6=\tfrac12$ is just a number, so the equation is $\tfrac12x+3y=0$.
Not linear (c) contains the product $xy$ of two variables. (d) contains $\sqrt x=x^{1/2}$, a power other than $1$. (e) contains $\tfrac1x=x^{-1}$.
Rule of thumb Ask what happens to the variables, not to the numbers: each variable must appear only as "constant $\times$ variable".
๐ Definition 1.2 โ System, solution, solution set
A system of linear equations (or linear system) is a finite collection of linear equations in the same variables $x_1,\dots,x_n$. An ordered list of numbers $(s_1,s_2,\dots,s_n)$ is a solution of the system if every equation becomes a true statement when we substitute $x_1=s_1,\ \dots,\ x_n=s_n$. The set of all solutions is the solution set. Two systems are equivalent if they have the same solution set.
A system with $m$ equations and $n$ unknowns is called an $m\times n$ system ("$m$ by $n$"). The cafรฉ problem is a $2\times2$ system. To solve a system means to describe its whole solution set โ not just to find one solution.
โ๏ธ Example 2 โ Checking solutions, and a solution set described by a parameter
(a) Is $(2,1,-1)$ a solution of the system $\ x+y+z=2,\ \ 2x-y+3z=0$? Is $(1,1,0)$?
(b) Describe all solutions of the single equation $2x-3y=6$.
๐ก Show solutionโถ
(a) First triple Substitute $x=2$, $y=1$, $z=-1$: $\ 2+1+(-1)=2$ โ and $2(2)-1+3(-1)=0$ โ. Both equations hold, so $(2,1,-1)$ is a solution.
(a) Second triple For $(1,1,0)$: $1+1+0=2$ โ but $2(1)-1+3(0)=1\ne0$ โ. It satisfies the first equation only, so it is not a solution of the system. A solution must satisfy every equation.
(b) Parameter One equation cannot pin down two unknowns, so we let one of them be free. Put $y=t$, where $t$ is any real number (a parameter). Then $2x=6+3t$, so $$x=3+\tfrac32t,\qquad y=t,\qquad t\in\mathbb{R}.$$
Check $2\left(3+\tfrac32t\right)-3t=6+3t-3t=6$ โ for every $t$. For instance $t=0$ gives $(3,0)$ and $t=2$ gives $(6,2)$. Geometrically, the solution set is the whole line $2x-3y=6$, and $t$ tells us where we are on it.
The Geometry of Solution Sets
An equation $ax+by=c$ with $a$ and $b$ not both zero is a straight line in the plane, and its solutions are exactly the points on the line. A solution of a $2\times2$ system is therefore a point lying on both lines. Two lines in the plane can be arranged in only three ways:
Lines crossThey meet in exactly one point โ exactly one solution.
Lines are parallel and distinctThey never meet โ no solution.
Lines coincideThey are the same line โ infinitely many solutions, one for every point on the line.
๐ Definition 1.3 โ Consistent and inconsistent
A linear system is consistent if it has at least one solution, and inconsistent if it has no solution.
Surprisingly, the same three outcomes are the only possibilities for every linear system, however many equations and unknowns it has. A system can never have exactly two solutions, or exactly five.
๐ Theorem 1.1 โ The number of solutions
Every system of linear equations has either no solution, exactly one solution, or infinitely many solutions.
Why. Suppose a system has two different solutions $p=(p_1,\dots,p_n)$ and $q=(q_1,\dots,q_n)$. For any real number $t$ consider the list $r=p+t(q-p)$, that is $r_i=p_i+t(q_i-p_i)$. Take any equation $a_1x_1+\cdots+a_nx_n=b$ of the system. Because $p$ and $q$ both satisfy it,
$$a_1r_1+\cdots+a_nr_n=\underbrace{(a_1p_1+\cdots+a_np_n)}_{=\,b}+t\big(\underbrace{(a_1q_1+\cdots+a_nq_n)}_{=\,b}-\underbrace{(a_1p_1+\cdots+a_np_n)}_{=\,b}\big)=b+t(b-b)=b.$$
So $r$ is a solution for every $t$ โ infinitely many solutions, all on the line through $p$ and $q$. Hence once there are two solutions there are infinitely many.
โ๏ธ Example 3 โ Three systems, three pictures
Solve each system and describe it geometrically.
(a) $\ x+y=4,\ \ x-y=2$ (b) $\ x+2y=4,\ \ 2x+4y=2$ (c) $\ x+2y=4,\ \ 3x+6y=12$
๐ก Show solutionโถ
(a) Adding the equations eliminates $y$: $2x=6$, so $x=3$, and then $y=4-3=1$. Exactly one solution, $(3,1)$: the lines (slopes $-1$ and $1$) cross at that point. Check: $3+1=4$ โ, $3-1=2$ โ.
(b) Subtract twice the first equation from the second: $(2x+4y)-2(x+2y)=2-8$, i.e. $0=-6$. This is false whatever $x$ and $y$ are, so the system is inconsistent. Both lines have slope $-\tfrac12$ but different $y$-intercepts ($2$ and $\tfrac12$): they are parallel and never meet.
(c) The second equation is $3$ times the first, so subtracting $3\times$(first) from it gives $0=0$, which is always true and tells us nothing new. Every solution of $x+2y=4$ works. With $y=t$: $$x=4-2t,\quad y=t,\qquad t\in\mathbb{R}.$$ The two equations describe the same line: infinitely many solutions.
Pattern A false statement such as $0=-6$ signals "no solution"; an identity $0=0$ signals a redundant equation. Interactive 1 lets you produce all three cases by changing the coefficients.
Interactive 1 โ Two Lines, Three Possible Answers
Each equation $ax+by=c$ is a line. Change the coefficients and watch the verdict. Press Play to let line โก swing round: as it turns parallel to line โ the crossing point races off to infinity โ and when the two lines lie on top of each other every point is a solution. The number $a_1b_2-a_2b_1$ is zero exactly when the lines are parallel or the same.
Line โ
Line โก
line โ โ
line โกโ
$a_1b_2-a_2b_1$โ
solution setโ
Three unknowns: planes
An equation $ax+by+cz=d$ in three variables (with $a,b,c$ not all zero) describes a plane in space. A $3\times3$ system asks for the points common to three planes. Theorem 1.1 still holds, but there are more ways for the planes to arrange themselves:
How the three planes sit
Solution set
Number of solutions
meet in a single point (like the floor and two walls of a room)
one point
exactly one
all pass through one common line (like pages of an open book)
a line
infinitely many
all three are the same plane
a plane
infinitely many
two (or three) are parallel and distinct
empty
none
each pair meets in a line, but the three lines are parallel (a "tent" or prism)
empty
none
With more than three unknowns we can no longer draw pictures, but the algebra of this chapter works in exactly the same way โ that is the power of linear algebra.
Matrices for Linear Systems
When we solve a system by elimination, the names of the variables just come along for the ride; all the real work is done on the coefficients and the constants. So we write those numbers, and only those, in a rectangular array.
๐ Definition 1.4 โ Coefficient matrix and augmented matrix
A matrix is a rectangular array of numbers; an $m\times n$ matrix has $m$ rows and $n$ columns, and its entry in row $i$ and column $j$ is written $a_{ij}$. For the $m\times n$ system
$$\begin{aligned}a_{11}x_1+a_{12}x_2+\cdots+a_{1n}x_n&=b_1\\ &\ \ \vdots\\ a_{m1}x_1+a_{m2}x_2+\cdots+a_{mn}x_n&=b_m\end{aligned}$$
the coefficient matrix $A$ is the $m\times n$ matrix of the coefficients $a_{ij}$, and the augmented matrix is $A$ with the column of constants $b_1,\dots,b_m$ attached on the right, separated by a vertical bar:
$$A=\begin{bmatrix}a_{11}&\cdots&a_{1n}\\ \vdots&&\vdots\\ a_{m1}&\cdots&a_{mn}\end{bmatrix},\qquad \left[\begin{array}{ccc|c}a_{11}&\cdots&a_{1n}&b_1\\ \vdots&&\vdots&\vdots\\ a_{m1}&\cdots&a_{mn}&b_m\end{array}\right].$$
Row $i$ of the augmented matrix is equation $i$; column $j$ (left of the bar) holds the coefficients of $x_j$.
โ๏ธ Example 4 โ From system to matrix and back
(a) Write the augmented matrix of the system
$$x+2y-z=-3,\qquad 2z+3y=1,\qquad 2x-y=11-4z.$$
(b) Write the system whose augmented matrix is $\left[\begin{array}{ccc|c}2&-1&0&3\\1&0&5&-2\end{array}\right]$, using variables $x_1,x_2,x_3$.
๐ก Show solutionโถ
(a) Tidy first Put every equation in the standard form with the variables in the same order $x,y,z$ on the left and the constant on the right: $$\begin{aligned}x+2y-z&=-3\\ 0x+3y+2z&=1\\ 2x-y+4z&=11\end{aligned}$$ The missing $x$ in the second equation gets coefficient $0$, and $-4z$ moves to the left as $+4z$.
(a) Matrix $$\left[\begin{array}{ccc|c}1&2&-1&-3\\0&3&2&1\\2&-1&4&11\end{array}\right]$$ (You can check that $(x,y,z)=(1,-1,2)$ satisfies all three equations.)
(b) Read each row as an equation, columns giving the coefficients of $x_1,x_2,x_3$: $$2x_1-x_2=3,\qquad x_1+5x_3=-2.$$ Zero entries simply mean the variable is absent.
Elementary Row Operations
When you solved Example 3 you instinctively did three kinds of thing: rearranged equations, multiplied an equation by a number, and added a multiple of one equation to another. None of these changes the solutions. In matrix language they become operations on rows.
The three elementary row operations on a matrix are:
Interchange two rows: $R_i\leftrightarrow R_j$.
Scale a row by a nonzero constant $c$: $R_i\to cR_i$ ($c\ne0$).
Replace a row by itself plus a multiple of a different row: $R_i\to R_i+cR_j$ ($j\ne i$).
Two matrices are row equivalent if one can be turned into the other by a finite sequence of elementary row operations.
๐ Theorem 1.2 โ Row operations do not change the solution set
If the augmented matrices of two linear systems are row equivalent, then the two systems have exactly the same solution set.
Why. It is enough to look at one operation. Any solution of the old system still satisfies the new one: swapping equations changes nothing; if $(\dots)=b_i$ holds then $c(\dots)=cb_i$ holds; and if equations $i$ and $j$ hold, so does their sum "equation $i$ $+$ $c\times$ equation $j$". Conversely, every operation can be undone by another elementary operation โ $R_i\leftrightarrow R_j$ by itself, $R_i\to cR_i$ by $R_i\to\tfrac1cR_i$ (this is why $c\ne0$ is required), and $R_i\to R_i+cR_j$ by $R_i\to R_i-cR_j$ โ so every solution of the new system is also a solution of the old one.
Geometrically, in a $2\times2$ system the replacement $R_2\to R_2+cR_1$ rotates line โก about the solution point: the new line still passes through every point common to both old lines. Watch it happen.
Interactive 2 โ Row Operations Move the Lines, Not the Solution
Choose a system and press Next operation (or Play). The highlighted row is the one being changed. A replacement $R_2\to R_2+cR_1$ swings line โก round the crossing point; scaling a row does not move its line at all; interchanging rows just swaps the colours. At the end the lines are as simple as possible โ and they still cross at the same point. For a system with no solution, line โก slides away to infinity and leaves the impossible equation $0=c$.
augmented matrix
stepโ
operationโ
common point(s)โ
โ ๏ธ Common mistakes with row operations
1. Never scale by zero. "$R_2\to0R_2$" wipes out an equation and cannot be undone, so it can create solutions that were not there before.
2. Only the named row changes. In $R_2\to R_2-2R_1$, row 1 is used but not altered. Copy it down unchanged.
3. Do the whole row โ including the right-hand side. Forgetting to apply the operation to the entry after the bar is the most common arithmetic slip.
4. "Replace a row by another row" is not allowed. $R_2\to R_1$ throws away equation 2. Replacement always adds a multiple of another row to the row being changed.
5. More unknowns than equations does not guarantee solutions. $x+y+z=1$ and $x+y+z=2$ is a $2\times3$ system with no solution (two parallel planes).
Triangular Systems and Back-Substitution
Some systems are easy to solve because they are already "staircase-shaped": the last equation involves only the last variable, the one above it only the last two, and so on. Such a system is in triangular form, and its augmented matrix has only zeros below the diagonal. We solve it from the bottom up.
Substitute that value into the equation above it and solve for the next variable.
Keep moving upwards, substituting every value already found, until all variables are known.
If a variable never gets its own equation (there are fewer equations than unknowns), make it a free parameter $t$ and express the others in terms of $t$.
(b) Free variable Two equations, three unknowns: $z$ has no equation of its own, so let $z=t$. Then $y=1-t$ and $x=3-2y+z=3-2(1-t)+t=1+3t$: $$x=1+3t,\quad y=1-t,\quad z=t,\qquad t\in\mathbb{R}.$$ Check: $(1+3t)+2(1-t)-t=3$ โ and $(1-t)+t=1$ โ. The two planes meet in a line โ infinitely many solutions.
โ๏ธ Example 6 โ Row operations, then back-substitution
Use row operations to bring the system to triangular form, then solve it:
$$x+y+z=6,\qquad 2x+3y+z=11,\qquad x-y+2z=5.$$
Clear column 1 $R_2\to R_2-2R_1$ and $R_3\to R_3-R_1$: $$\left[\begin{array}{ccc|c}1&1&1&6\\0&1&-1&-1\\0&-2&1&-1\end{array}\right]$$
Clear column 2 $R_3\to R_3+2R_2$: $$\left[\begin{array}{ccc|c}1&1&1&6\\0&1&-1&-1\\0&0&-1&-3\end{array}\right]$$ The matrix is now triangular.
Back-substitute $-z=-3$, so $z=3$; $y-z=-1$, so $y=2$; $x+y+z=6$, so $x=1$. Solution $(1,2,3)$.
Check In the original equations: $1+2+3=6$ โ, $2+6+3=11$ โ, $1-2+6=5$ โ. By Theorem 1.2 the triangular system and the original have the same solution set, so $(1,2,3)$ is the only solution. Section 1.2 turns this procedure into a complete algorithm โ Gaussian elimination โ that works for every system.
Exercises
I
Linear equations and solutions.
1 Is $4x-7y+z=0$ a linear equation?
2 Is $x^2+y=3$ a linear equation?
3 Is $\dfrac{x}{3}-\sqrt5\,y=1$ a linear equation?
4 Is $\dfrac1x+2y=7$ a linear equation?
5 Is $(1,-2,3)$ a solution of the system $\ 2x-y+z=7,\ \ x+y+z=2,\ \ x-3y-z=4$?
6 Which of $(0,1,2)$ and $(2,0,1)$ are solutions of $\ x+2y-z=0,\ \ 3x-y+2z=3$?
7 Describe all solutions of $3x-6y=9$ using a parameter.
8 Describe all solutions of $x_1+4x_2-2x_3=5$ using two parameters.
๐ก Show solutionsโถ
1 Yes: coefficients $4,-7,1$, constant term $0$.
2 No: $x^2$ is a second power of a variable.
3 Yes: it is $\tfrac13x-\sqrt5\,y=1$; $\tfrac13$ and $-\sqrt5$ are constants.
4 No: $\tfrac1x=x^{-1}$ is not a first power.
5 Yes: $2+2+3=7$ โ, $1-2+3=2$ โ, $1+6-3=4$ โ โ all three equations hold.
6 $(0,1,2)$: $0+2-2=0$ โ and $0-1+4=3$ โ, so it is a solution. $(2,0,1)$: $2+0-1=1\ne0$ โ, so it is not (no need to test the second equation).
7 Let $y=t$; then $3x=9+6t$, so $x=3+2t$, $y=t$ ($t\in\mathbb{R}$). Check: $3(3+2t)-6t=9$ โ.
8 Let $x_2=s$, $x_3=t$; then $x_1=5-4s+2t$. Solution set: $(5-4s+2t,\ s,\ t)$ with $s,t\in\mathbb{R}$ โ a plane in $\mathbb{R}^3$.
II
Systems of two equations in two unknowns. Solve each system, say whether it is consistent, and describe the two lines.
1 $x+y=5,\ \ x-y=1$
2 $2x+3y=7,\ \ x-y=1$
3 $3x-y=4,\ \ 6x-2y=1$
4 $x-2y=3,\ \ -2x+4y=-6$
5 $4x+y=2,\ \ x+2y=-3$
6 $\dfrac x2+\dfrac y3=1,\ \ x-y=7$
7 For which values of $k$ does $\ x+ky=2,\ \ 3x+6y=5\ $ have (a) exactly one solution, (b) no solution, (c) infinitely many solutions?
8 For which values of $a$ and $b$ does $\ 2x-y=a,\ \ 6x-3y=b\ $ have infinitely many solutions? Can it ever have exactly one?
Check: draw the two lines of parts 1โ6 and see where (or whether) they meet
line โ โ
line โกโ
answerโ
๐ก Show solutionsโถ
1 Add: $2x=6$, $x=3$; then $y=2$. One solution $(3,2)$; consistent; the lines cross. Check: $3+2=5$, $3-2=1$ โ.
2 From the second equation $x=1+y$; then $2(1+y)+3y=7$, $5y=5$, $y=1$, $x=2$. One solution $(2,1)$; the lines cross.
3 $R_2\to R_2-2R_1$ gives $0=1-8=-7$: false. No solution; inconsistent. Both lines have slope $3$ but different intercepts ($-4$ and $-\tfrac12$): parallel.
4 The second equation is $-2$ times the first, so $R_2\to R_2+2R_1$ gives $0=0$. Infinitely many solutions: $x=3+2t$, $y=t$ ($t\in\mathbb{R}$); consistent; the two equations describe the same line.
5 From the first, $y=2-4x$; then $x+2(2-4x)=-3$, $-7x=-7$, $x=1$, $y=-2$. One solution $(1,-2)$. Check: $4-2=2$ โ, $1-4=-3$ โ.
6 Multiply the first equation by $6$: $3x+2y=6$. With $x=7+y$: $21+3y+2y=6$, $y=-3$, $x=4$. One solution $(4,-3)$. Check: $\tfrac42+\tfrac{-3}{3}=2-1=1$ โ.
7 $a_1b_2-a_2b_1=1\cdot6-3k=6-3k$, which is zero only for $k=2$. (a) For $k\ne2$ the lines have different slopes: exactly one solution. (b) For $k=2$: $R_2\to R_2-3R_1$ gives $0=5-6=-1$, false โ no solution (parallel lines). (c) Never: when $k=2$ the right-hand sides do not match ($3\times2=6\ne5$).
8 The left side of the second equation is always $3$ times the first, so the lines are always parallel or identical โ never exactly one solution. $R_2\to R_2-3R_1$ gives $0=b-3a$. Infinitely many solutions exactly when $b=3a$ (then $x=t$, $y=2t-a$); no solution when $b\ne3a$.
III
Augmented matrices and row operations.
1 Write the augmented matrix of $\ 3x-2y=5,\ \ x+4y=-1$.
2 Write the coefficient matrix and the augmented matrix of $\ x_1+x_3=4,\ \ 2x_2-x_3=0,\ \ x_1-x_2+3x_3=-2$.
3 Write the system with augmented matrix $\left[\begin{array}{ccc|c}1&-3&0&2\\0&1&4&5\\0&0&1&1\end{array}\right]$ (variables $x,y,z$) and solve it by back-substitution.
4 Apply $R_2\to R_2-3R_1$ to $\left[\begin{array}{cc|c}1&2&4\\3&5&9\end{array}\right]$, then solve the system.
5 Apply, in order, $R_1\leftrightarrow R_2$, $\ R_2\to R_2-2R_1$, $\ R_2\to\tfrac15R_2$ to $\left[\begin{array}{cc|c}2&1&3\\1&-2&4\end{array}\right]$, writing the matrix after each step. Then solve.
6 Which of the following are elementary row operations? (a) $R_2\to R_2+5R_1$; (b) $R_1\to0R_1$; (c) $R_2\to R_1$; (d) $R_1\leftrightarrow R_3$; (e) $R_3\to-\tfrac12R_3$.
7 Give the elementary row operation that undoes each of: (a) $R_2\to R_2-2R_1$; (b) $R_2\to4R_2$; (c) $R_1\leftrightarrow R_3$.
8 Show that $\left[\begin{array}{cc|c}1&2&3\\2&4&7\end{array}\right]$ and $\left[\begin{array}{cc|c}1&2&3\\0&0&1\end{array}\right]$ are row equivalent. What does the second matrix tell you about the system?
4 $R_2\to R_2-3R_1$: $\left[\begin{array}{cc|c}1&2&4\\0&-1&-3\end{array}\right]$. So $-y=-3$, $y=3$, and $x=4-2(3)=-2$. Solution $(-2,3)$. Check in the original: $-2+6=4$ โ, $-6+15=9$ โ.
6 (a) yes (replacement); (b) no โ scaling must use a nonzero constant; (c) no โ it replaces row 2 by row 1 and loses equation 2; (d) yes (interchange); (e) yes (scaling by $-\tfrac12\ne0$).
8 $R_2\to R_2-2R_1$ turns $(2,4,7)$ into $(0,0,1)$, so the matrices are row equivalent. The second row means $0x+0y=1$, i.e. $0=1$: the system is inconsistent โ by Theorem 1.2 so is the original (its lines $x+2y=3$ and $2x+4y=7$ are parallel).
IV
Three unknowns, applications and ideas.
1 Solve by back-substitution: $\ x+y-z=2,\ \ y+3z=11,\ \ 2z=6$.
2 Solve by back-substitution: $\ x-y+2z=4,\ \ y-z=1$.
3 Use row operations to reach triangular form, then solve: $\ x+2y+z=8,\ \ 2x+y-z=1,\ \ x-y+2z=5$.
4 Show that $\ x+y+z=2,\ \ x-y+z=0,\ \ 2x+2z=3\ $ is inconsistent, and describe the three planes.
5Ticket sales. A cinema sold $450$ tickets for one show. Adult tickets cost โฌ9 and child tickets โฌ5, and the takings were โฌ3410. How many of each were sold?
6Pharmacy. A pharmacist mixes a $10\%$ saline solution with a $30\%$ saline solution to make $50$ mL of an $18\%$ solution. How much of each is needed?
7Finance. An investor places $\$20\,000$ in three funds paying $2\%$, $4\%$ and $5\%$ per year, earning $\$680$ in total interest in one year. She invested twice as much in the $2\%$ fund as in the $5\%$ fund. Set up a $3\times3$ system and find each amount.
8Challenge. For which values of $h$ and $k$ does the system $\ x+y+z=1,\ \ x+2y+3z=2,\ \ x+3y+hz=k\ $ have (a) exactly one solution, (b) no solution, (c) infinitely many solutions? Give the solution set in case (c).
9Conceptual. Explain why a system of linear equations can never have exactly two solutions.
10Conceptual. Give (a) a system of three equations in two unknowns with exactly one solution, and (b) a system of two equations in three unknowns with no solution.
2 $z$ is free: $z=t$. Then $y=1+t$ and $x=4+y-2z=4+(1+t)-2t=5-t$. Solution set $(5-t,\ 1+t,\ t)$, $t\in\mathbb{R}$ โ a line of solutions. Check: $(5-t)-(1+t)+2t=4$ โ.
4 $R_2\to R_2-R_1$ gives $(0,-2,0\,|\,-2)$; $R_3\to R_3-2R_1$ gives $(0,-2,0\,|\,-1)$; then $R_3\to R_3-R_2$ gives $(0,0,0\,|\,1)$, i.e. $0=1$. Inconsistent. No two of the planes are parallel (their coefficient lists are not multiples of each other), so each pair meets in a line; but adding the first two equations gives $2x+2z=2$, which contradicts $2x+2z=3$ โ the three lines of intersection are parallel, forming a "tent" with no common point.
5 Let $a$ = adult and $c$ = child tickets: $a+c=450$, $9a+5c=3410$. Substituting $c=450-a$: $9a+2250-5a=3410$, $4a=1160$, $a=290$, $c=160$. So $290$ adult and $160$ child tickets. Check: $9(290)+5(160)=2610+800=3410$ โ.
6 Let $x$ mL of $10\%$ and $y$ mL of $30\%$: $x+y=50$ and $0.10x+0.30y=0.18(50)=9$. Multiply the second by $10$: $x+3y=90$. Subtract the first: $2y=40$, $y=20$, $x=30$. So $30$ mL of the $10\%$ and $20$ mL of the $30\%$ solution. Check: $3+6=9$ mL of salt โ.
7 Let $x,y,z$ be the dollars in the $2\%$, $4\%$, $5\%$ funds: $$x+y+z=20\,000,\qquad 0.02x+0.04y+0.05z=680,\qquad x-2z=0.$$ With $x=2z$: $y=20\,000-3z$, and $0.04z+0.04(20\,000-3z)+0.05z=680$, i.e. $800-0.03z=680$, $z=4000$. So $x=\$8000$, $y=\$8000$, $z=\$4000$. Check: $160+320+200=680$ โ.
8 $R_2\to R_2-R_1$: $(0,1,2\,|\,1)$; $R_3\to R_3-R_1$: $(0,2,h-1\,|\,k-1)$; $R_3\to R_3-2R_2$:$(0,0,h-5\,|\,k-3)$. (a) $h\ne5$: then $z=\tfrac{k-3}{h-5}$ and back-substitution gives exactly one solution. (b) $h=5$, $k\ne3$: the last row says $0=k-3\ne0$ โ no solution. (c) $h=5$, $k=3$: the last row is $0=0$; with $z=t$, $y=1-2t$, $x=1-y-z=t$, so the solution set is $(t,\ 1-2t,\ t)$, $t\in\mathbb{R}$.
9 By Theorem 1.1: if $p$ and $q$ are two different solutions, then $p+t(q-p)$ is a solution for every real $t$, and different values of $t$ give different points. So two solutions force infinitely many.
10 (a) For example $x+y=2$, $x-y=0$, $2x+y=3$: the first two give $(1,1)$, and $2(1)+1=3$ โ, so $(1,1)$ is the only solution โ three lines through one point. (b) For example $x+y+z=1$, $x+y+z=3$: two parallel planes, no common point.
๐ฌ ResearchโTeaching NexusResearch-tutored ยท optional, about 30 minutes
Research question. How do animal-feed producers blend ingredients so that a feed meets exact nutritional targets โ and when is a target impossible?
Why it matters. Poultry and fish farms around the world buy feed mixed to precise specifications for protein and energy. Too little protein slows growth; too much wastes the most expensive ingredient and pollutes water with nitrogen. Every feed formula starts as a linear system: one equation per nutrient, one unknown per ingredient.
What to do.
Sample case (typical illustrative values): per kg, maize supplies $0.09$ kg protein and $3.3$ Mcal energy; soybean meal $0.44$ kg and $2.2$ Mcal; wheat bran $0.15$ kg and $1.3$ Mcal. A farm needs $100$ kg of feed with $20.4$ kg protein and $267$ Mcal energy. Let $x,y,z$ be the kg of maize, soybean meal and bran.
Write the $3\times3$ system (mass, protein, energy) and its augmented matrix. Use row operations to reach triangular form and back-substitute.
Check your answer in all three original equations.
Now ask for $30$ kg protein with the same mass and energy. Solve again. Is the answer physically possible? Explain.
Find, as an inequality, the range of protein targets (with $100$ kg and $267$ Mcal fixed) for which all three amounts are non-negative.
Tools: Pen and paper; a spreadsheet or any matrix calculator to confirm.
Hand in: A half-page note with both systems, both solutions, your feasible protein range, and one sentence on what a nutritionist should do when the system's only solution has a negative amount.
Going further. Add a fourth ingredient (for example fish meal, about $0.60$ kg protein and $2.8$ Mcal per kg). Now there are more unknowns than equations: show that there are infinitely many blends, and look up how linear programming chooses the cheapest one.
Instructor note: the first system has the unique solution $x=55$, $y=30$, $z=15$ kg. With $30$ kg protein the solution is $x=\tfrac{13115}{317}\approx41.37$, $y=\tfrac{19110}{317}\approx60.28$, $z=-\tfrac{525}{317}\approx-1.66$ kg โ a negative amount of bran, so the target is impossible with these ingredients. In general $x=\tfrac{26615-450P}{317}$, $y=\tfrac{1000P-10890}{317}$, $z=\tfrac{15975-550P}{317}$, so all three are non-negative exactly when $10.89\le P\le\tfrac{639}{22}\approx29.05$ kg.
๐งช Self-check quiz
0/5 correct
Q1. Which equation is linear?
Q2. A $2\times2$ system whose two lines are parallel and distinct has
Q3. With variables in the order $x,y,z$, the augmented matrix of $\ x-y=3,\ \ 2x+z=1\ $ is
Q4. Which of these is not an elementary row operation?
Q5. Back-substitution in $\ x+2y=7,\ \ 3y=6\ $ gives